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1 2/6/12 Overview CS 112 Introduction to Programming What is recursion? When one function calls itself directly or indirectly. (Spring 2012) Why learn recursion? New mode of thinking. Powerful programming paradigm. Lecture #13: Recursion Zhong Shao Many computations are naturally self-referential. Mergesort, FFT, gcd, depth-first search. Linked data structures. A folder contains files and other folders. Department of Computer Science Yale University Office: 314 Watson Closely related to mathematical induction. Reproductive Parts M. C. Escher, 1948 Acknowledgements: some slides used in this class are taken directly or adapted from those accompanying the tetbook: Introduction to Programming in Java: An Interdisciplinary Approach by Robert Sedgewick and Kevin Wayne (Copyright ) 2 Greatest Common Divisor Greatest Common Divisor Gcd. Find largest integer that evenly divides into p and q. Gcd. Find largest integer d that evenly divides into p and q. E. gcd(4032, 1272) = 24. Euclid's algorithm. [Euclid 300 BCE] 4032 = = gcd = = 24 "p if q = 0 gcd( p, q) = # $ gcd(q, p q) otherwise base case reduction step, converges to base case! Applications. Simplify fractions: 1272/4032 = 53/168. RSA cryptosystem. gcd(4032, 1272) 3 = = = = = gcd(1272, 216) gcd(216, 192) gcd(192, 24) gcd(24, 0) =

2 2/6/12 Greatest Common Divisor Greatest Common Divisor Gcd. Find largest integer d that evenly divides into p and q. base case "p if q = 0 gcd( p, q) = # $ gcd(q, p q) otherwise! "p if q = 0 gcd( p, q) = # $ gcd(q, p q) otherwise reduction step, converges to base case reduction step, converges to base case Java implementation. q base case! p q Gcd. Find largest integer d that evenly divides into p and q. p q p = 8 q = 3 gcd(p, q) = public static int gcd(int p, int q) { if (q == 0) return p; else return gcd(q, p q); base case reduction step gcd 5 6 Recursive Graphics 8 2

3 Htree H-tree of order n. Draw an H. and half the size Recursively draw 4 H-trees of order n-1, one connected to each tip. tip size order 1 order 2 order Htree in Java Animated H-tree public class Htree { public static void draw(int n, double sz, double, double y) { if (n == 0) return; double 0 = - sz/2, 1 = + sz/2; double y0 = y - sz/2, y1 = y + sz/2; Animated H-tree. Pause for 1 second after drawing each H. StdDraw.line(0, y, 1, y); StdDraw.line(0, y0, 0, y1); StdDraw.line(1, y0, 1, y1); draw the H, centered on (, y) draw(n-1, sz/2, 0, y0); draw(n-1, sz/2, 0, y1); draw(n-1, sz/2, 1, y0); draw(n-1, sz/2, 1, y1); recursively draw 4 half-size Hs public static void main(string[] args) { int n = Integer.parseInt(args[0]); draw(n,.5,.5,.5); 11 12

4 Towers of Hanoi Towers of Hanoi Move all the discs from the leftmost peg to the rightmost one. Only one disc may be moved at a time. A disc can be placed either on empty peg or on top of a larger disc. start finish Towers of Hanoi demo Edouard Lucas (1883) 14 Towers of Hanoi Legend Towers of Hanoi: Recursive Solution Q. Is world going to end (according to legend)? 64 golden discs on 3 diamond pegs. World ends when certain group of monks accomplish task. Q. Will computer algorithms help? Move n-1 smallest discs right. Move largest disc left. cyclic wrap-around Move n-1 smallest discs right

5 Towers of Hanoi: Recursive Solution Towers of Hanoi: Recursive Solution public class TowersOfHanoi { public static void moves(int n, boolean left) { if (n == 0) return; moves(n-1,!left); if (left) System.out.println(n + " left"); else System.out.println(n + " right"); moves(n-1,!left); public static void main(string[] args) { int N = Integer.parseInt(args[0]); moves(n, true); java TowersOfHanoi 3 1 left 2 right 1 left 3 left 1 left 2 right 1 left every other move is smallest disc java TowersOfHanoi 4 2 left 3 right 2 left 4 left 2 left 3 right 2 left moves(n, true) : move discs 1 to n one pole to the left moves(n, false): move discs 1 to n one pole to the right subdivisions of ruler smallest disc Towers of Hanoi: Recursion Tree Towers of Hanoi: Properties of Solution 1 14 n, left 3, true Remarkable properties of recursive solution. Takes 2 n - 1 moves to solve n disc problem. Sequence of discs is same as subdivisions of ruler. Every other move involves smallest disc , false 1, true 1, true 2, false , true 1, true Recursive algorithm yields non-recursive solution! Alternate between two moves: move smallest disc to right if n is even make only legal move not involving smallest disc to left if n is odd Recursive algorithm may reveal fate of world. Takes 585 billion years for n = 64 (at rate of 1 disc per second). Reassuring fact: any solution takes at least this long! 1 left 2 right 1 left 3 left 1 left 2 right 1 left 19 20

6 2/6/12 Divide-and-Conquer Fibonacci Numbers Divide-and-conquer paradigm. Break up problem into smaller subproblems of same structure. Solve subproblems recursively using same method. Combine results to produce solution to original problem. Divide et impera. Veni, vidi, vici. - Julius Caesar Many important problems succumb to divide-and-conquer. FFT for signal processing. Parsers for programming languages. Multigrid methods for solving PDEs. Quicksort and mergesort for sorting. Hilbert curve for domain decomposition. Quad-tree for efficient N-body simulation. Midpoint displacement method for fractional Brownian motion. 21 Fibonacci Numbers Fibonacci Numbers and Nature Fibonacci numbers. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, Fibonacci numbers. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, #0 if n = 0 F(n) = $ 1 if n = 1 & F(n "1) + F(n " 2) otherwise #0 if n = 0 F(n) = $ 1 if n = 1 & F(n "1) + F(n " 2) otherwise!! L. P. Fibonacci ( ) pinecone Fibonacci rabbits 23 cauliflower 24 6

7 A Possible Pitfall With Recursion Recursion Challenge 1 (difficult but important) Fibonacci numbers. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, Q. Is this an efficient way to compute F(50)? # 0 if n = 0 F(n) = $ 1 if n =1 & F(n "1) + F(n " 2) otherwise public static long F(int n) { if (n == 0) return 0; if (n == 1) return 1; return F(n-1) + F(n-2); A natural for recursion? public static long F(int n) { if (n == 0) return 0; if (n == 1) return 1; return F(n-1) + F(n-2); A. No, no, no! This code is spectacularly inefficient. F(50) F(49) F(48) F(50) is called once. F(49) is called once. F(48) is called 2 times. F(48) F(47) F(47) F(46) F(47) F(46) F(46) F(45) F(46) F(45) F(45) F(44) F(47) is called 3 times. F(46) is called 5 times. F(45) is called 8 times.... recursion tree for naïve Fibonacci function F(1) is called 12,586,269,025 times. 25 F(50) 26 Recursion Challenge 2 (easy and also important) Summary Q. Is this a more efficient way to compute F(50)? public static long F(int n) { long[] F = new long[n+1]; F[0] = 0; F[1] = 1; for(int i = 2; i <= n; i++) F[i] = F[i-1] + F[i-2]; return F[n]; FYI: classic math F(n) = " n # (1#") n 5 = $ " n 5 φ = golden ratio How to write simple recursive programs? Base case, reduction step. Trace the eecution of a recursive program. Use pictures. Why learn recursion? New mode of thinking. Powerful programming tool. Towers of Hanoi by W. A. Schloss. A. Yes. This code does it with 50 additions. Lesson. Don t use recursion to engage in eponential waste. Divide-and-conquer. Elegant solution to many important problems. Contet. This is a special case of an important programming technique known as dynamic programming (stay tuned)

8 Collatz Sequence Etra Slides Collatz sequence. If n is 1, stop. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. E public static void collatz(int n) {! System.out.print(n + " ");! if (n == 1) return;! if (n 2 == 0) collatz(n / 2);! collatz(3*n + 1);! 30 Fractional Brownian Motion Fractional Brownian Motion Physical process which models many natural and artificial phenomenon. Price of stocks. Dispersion of ink flowing in water. Rugged shapes of mountains and clouds. Fractal landscapes and tetures for computer graphics. 32

9 Simulating Brownian Motion Simulating Brownian Motion: Java Implementation Midpoint displacement method. Maintain an interval with endpoints ( 0, y 0 ) and ( 1, y 1 ). Divide the interval in half. Choose δ at random from Gaussian distribution. Set m = ( )/2 and y m = (y 0 + y 1 )/2 + δ. Recur on the left and right intervals. Midpoint displacement method. Maintain an interval with endpoints ( 0, y 0 ) and ( 1, y 1 ). Divide the interval in half. Choose δ at random from Gaussian distribution. Set m = ( )/2 and y m = (y 0 + y 1 )/2 + δ. Recur on the left and right intervals. public static void curve(double 0, double y0, double 1, double y1, double var) { if (1-0 < 0.01) { StdDraw.line(0, y0, 1, y1); return; double m = (0 + 1) / 2; double ym = (y0 + y1) / 2; ym += StdRandom.gaussian(0, Math.sqrt(var)); curve(0, y0, m, ym, var/2); curve(m, ym, 1, y1, var/2); variance halves at each level; change factor to get different shapes Plasma Cloud Plasma Cloud Plasma cloud centered at (, y) of size s. Each corner labeled with some grayscale value. Divide square into four quadrants. The grayscale of each new corner is the average of others. center: average of the four corners + random displacement others: average of two original corners Recur on the four quadrants. c 1 +c 2 c 1 2 c 2 c 1 +c 3 2 c 2 +c 4 2 (c 1 +c 2 +c 3 +c 4 ) 4 + " c 3 c 3 +c 4 c

10 Brownian Landscape Reference: 37

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